Two identical 0.50-kg carts, each 0.10 m long, are at rest on a low-friction track and are connected by a spring that is initially at its relaxed length of 0.50 m and is of negligible inertia. You give the cart on the left a push to the right (that is, toward the other cart), exerting a constant 6.0-N force. You stop pushing at the instant when the cart has moved 0.60m . At this instant, the relative velocity of the two carts is zero and the spring is compressed to a length of 0.30 m. A locking mechanism keeps the spring compressed, and the two carts continue moving to the right. By what amount do you change the system's kinetic energy?

The system's kinetic energy changes by 3.6 J

Explanation:

The given parameters are;

The number of cart = 2

The mass of each cart = 0.5kg

The initial length of the spring = 0.50 m

The final length of the spring =T0.3 m

The change in position of the first cart = 0.6 m

The energy given to the first cart = Work done by the force = Force × Displacement

The initial kinetic energy of the two cart moving together = Energy given to the first cart = 6.0 × 0.2 = 1.2J

The kinetic energy given to the two cart combined = The applied force × The total displacement of the two cart as they move together

The kinetic energy given to the two cart combined = 6.0 × (0.6 - 0.2)

The kinetic energy given to the two cart combined = 6.0 × 0.4 = 2.4 J

The total kinetic energy given to the two cart = 1.2 + 2.4 = 3.6 J

The total kinetic energy given to the two cart = 3.6 J

The system's kinetic energy changes by 3.6 J.

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